Problem: Find $\dfrac{d}{dx}\left(\dfrac1x\cos(x)\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{x\sin(x)-\cos(x)}{x^2}$ (Choice B) B $-\sin(x)-\dfrac1{x^2}$ (Choice C) C $\dfrac{-x\sin(x)-\cos(x)}{x^2}$ (Choice D) D $\dfrac{\sin(x)}{x^2}$
Answer: $\dfrac1x\cos(x)$ is the product of two, more basic, expressions: $\dfrac1x$ and $\cos(x)$. Therefore, the derivative of the expression can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\dfrac1x\cos(x)\right) \\\\ &=\dfrac{d}{dx}\left(\dfrac1x\right)\cos(x)+\dfrac1x\dfrac{d}{dx}(\cos(x))&&\gray{\text{The product rule}} \\\\ &=\dfrac{d}{dx}\left(x^{-1}\right)\cos(x)+\dfrac1x\dfrac{d}{dx}(\cos(x))&&\gray{\text{Write }\dfrac1x\text{ as a power}} \\\\ &=-1x^{-2}\cdot \cos(x)+\dfrac1x\cdot (-\sin(x))&&\gray{\text{Differentiate }x^{-1}\text{ and }\cos(x)} \\\\ &=-\dfrac1{x^{2}}\cdot \cos(x)+\dfrac1x\cdot (-\sin(x))&&\gray{\text{Write as positive exponents}} \\\\ &=\dfrac{-\cos(x)-x\sin(x)}{x^2}&&\gray{\text{Common denominator}} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left(\dfrac1x\cos(x)\right)=\dfrac{-\cos(x)-x\sin(x)}{x^2}$